4.1 Extreme Values of functions
Absolute Maximum – let f be a function with domain D, the point on the domain, c, in
Which f(x) ≤ f(c) for all x in D
Absolute Minimum – the point on the domain, c, when f(x) ≥ f(c) for all x in D
** Also called Extrema, Global Extrema, Extreme Values
Endpoint Extrema - extremes that occur at the endpoints.
Ex.’s
The Extreme Value Theorem – If f is continuous on closed interval [a.b] then f attains
both an an absolute maximum value M, and an absolute minimum value m in [a,b]. That
is, there are numbers x1 and x2 in [a,b] with f(x1) = m, and f(x2) = M and m≤f(x)≤M for
every other x in [a,b]
Ex.
Key Ingredients of Extreme Value Theorem:
1. Interval must be closed and finite
2. Function must be continuous
Local/Relative Maximum – an interior point c, of a functions domain if f(x)≤f(c) for all
x in some open interval containing c **occurs on hill of graph
Local/Relative Minimum – an interior point c, of a functions domain if f(x)≥f(c) for all
x in some open interval containing c **occurs on valley of graph
(largest and smallest in neighborhood)
** an absolute maximum/absolute minimum – can also be local max/min
Ex. – identify extrema
----------------------------------------------------------------------------------------Finding Extrema: **There is always a horizontal tangent line at the local extrema **
The First Derivative Theorem for Local Extreme Values:
If f has a local max. or min. value at an interior point, c, of its domain and its f′ is defined
at c, then f′(c) = 0
** basically says that a functions 1st derivative is always 0 at an interior point where the
function has a local extreme value and the derivative is defined—therefore the only
places where a function can possibly have an extreme value are:
1. Interior points where f′ = 0
2. Interior points where f′ is undefined
3. Endpoints of the domain of f
Critical Point – an interior point in the domain of a function f, where f′= 0 or is
undefined
a) relative extrema only occur at critical points – but a critical point doesn’t
have to be a relative extrema
**Therefore, the only domain points where a function can assume extreme values
are critical points and endpoints**
Steps to finding the Absolute Extrema of a continuous function on a finite,closed interval
#1. Evaluate f at all critical points ( ), and endpoints [ ] in the interval
#2. Take the largest and smallest of these values
Ex. Find the absolute max and min. values of each function on the given interval. Then
graph, identify points on graph where extrema occur.
F(x) =
4.2 The Mean Value Theorem/Rolle Theorem
** previous section we did Extreme Value Theorem – which says a continuous function
on a closed interval [a,b] must have both a minimum and
a maximum in the interval – both values can occur at the
Endpoints.
Rolle’s Theorem – gives conditions that guarantee the existence of an extreme value in
the interior of a closed interval.
Says: let f be continuous on the closed interval [a,b] and differentiable on the
Open interval (a,b). If f(a) = f(b) then there is at least 1 number, c, in
(a,b) such that f′(c)=0
** If above theorem is true --- its says that there must be at least 1 x value
Between a and b at which the graph has a horizontal tangent
Ex. Let f(x) = x4 -2x² Find all values of c in the interval (-2,2) such that f′(c)=0
Steps: 1. See if satisfies Rolle’s theorem by plugging in f(-2) and see if =f(2)
Step 2: If f(a)=f(b) then take the derivative and set = 0 to find value
Mean Value Theorem – says that if f is continuous on the closed interval [a,b] and
Differentiable on the open interval (a,b) then there exists a number,c
In (a,b) such that:
f′(c) = f(b) – f(a)
b–a
“Mean” – refers to the mean(average) rate of change of f in interval [a,b]
a) Geometrically – the theorem guarantees the existence of a tangent line that is
parallel to the secant line through pts (a,f(a)) and (b,f(b))
b) In terms of rate of change – implies that there must be a point in the open
interval (a,b) at which the instantaneous rate of change is equal
to the average rate of change over the interval [a,b]
Ex.
f′(c) = f(b) – f(a)
b–a
is the average change over f and f′(c) is instantaneous change --- the
mean value theorem says that at some interior point, the average rate of change must =
the instantaneous rate of change:
Ex. 2 stationary patrol cars equipped with radar are 5 miles apart on a highway. As a
truck passes the 1st patrol car, its speed is clocked at 55 mph. Four minutes later, when
the truck passes the 2nd patrol car, its speed is clocked at 50 mph. Prove that the truck
must have exceeded the speed limit (of 55 mph) at some time during the 4 minutes.
Solution:
Mathematical Consequences:
#1. Corollary 1: Functions with zero derivatives are constant
: if f′(x) = 0 at each point x of open interval (a,b) then f(x) = C for all x in(a,b)
where C is constant
#2. Corollary 2: Functions with the Same derivative differ by a constant
: if f′(x) = g′(x) at each point x in the open interval (a,b), then there exists a
constant C, such that f(x) = g(x) + C for all x in (a,b). That is f-g is constant
on (a,b)
Ex. Find the function f(x) where the derivative is sin x and whose graph passes through
the point (0,2)
Steps: 1. Work backwards
2. Then plug in x and y
3. Solve for C
4.3 Increasing and Decreasing Functions
Increasing Function – a function f, is increasing on an interval if for any 2 #’s, x1 and x2
in the interval, x1 < x2 implies f(x1) < f(x2)
**as x moves to the right that graph moves up
Decreasing Function – a function f, is decreasing on an interval if for any 2 #’s x1 and x2
in the interval x1 < x2 implies f(x1) > f(x2)
**as x moves to the right the graph moves down
** Increasing and Decreasing Functions must be satisfied for every pair of points in
the interval **
1. Positive derivative – implies the function is increasing
2. Negative derivative – implies the function is decreasing
3. Zero derivative – implies the function is constant on the interval
Look at figure
Theorem 3.5 Test for Increasing/Decreasing Functions:
Let f be a function that is continuous on the closed interval [a,b] and differentiable on the
open interval (a,b) then:
1. If f′(x) >0 for all x in (a,b) then f is increasing on [a,b]
2. If f′(x)<0 for all x in (a,b) the f is decreasing on [a,b]
3. If f′(x) =0 for all x in (a,b) then f is constant on [a,b]
Guidelines for finding intervals on which a function is increasing or decreasing:
1. Locate the critical #’s of f in (a,b) and use those #’s for test intervals
2. Determine the sign of f′(x) at one test value in each interval
3. Determine whether increasing or decreasing (theorem 3.5 on each interval)
Strictly Monotonic – a function that is either increasing or decreasing on the entire
interval
Ex. f(x) = x³ -- is strictly monotonic b/c it is increasing on the entire real line
Ex. Find the open interval on which f(x) = x³ -3/2x² is increasing or decreasing
First Derivative Test – used after you have determined the intervals on which a function
is increasing or decreasing to find the relative extrema.
Theorem 3.6 The First Derivative Test
Let c be an interval # of a f(x) that is continuous on an open interval I containing c. If f is
differentiable on the interval except possibly at c, then f(c) can be classified as:
1. If f′(x) changes from – to + at c, then f(c) is a relative minimum of f.
2. If f′(x) changes from + to – at c, then f(c) is a relative maximum of f
3. If f′(x) does not change sign at c, then f(c) is neither a relative max. nor min.
** Sign changes of f´(x) can only occur at critical points
**from above example f(x) = x³ - 3/2x² has a relative maximum at (0,0) b/c
Increasing immediately to left and decreasing immediately to right of x=0
**from above it has a relative minimum at (1,-½) b/c decreasing immediately to
Left of x=1 and increasing to the right of x=1
4.3 Concavity and Curve Sketching
So far we have found:
a) In section 4.3 we saw how first derivative test—tells us where a function is
increasing and decreasing
b) At the critical point – the 1st derivative test tells us whether there is a local max. or
local min.
In this section: we are told as to how the 2nd derivative gives information about the way
the graph of a differentiable function bends or turns.
The graph of a differentiable function y=f(x) is:
Concave Up – on an open interval I if f′(x) is increasing on I – graph will lie above its
tangents (slopes increasing)
Concave Down – on an open interval I if f′(x) is decreasing on I – graph lies below its
tangents (slopes decreasing)
The Second Derivative Test for Concavity:
Let y = f(x) be twice differentiable on interval I:
1. If f′′>0 on I, the graph of f over I is concave up
2. If f′′<0 on I, the graph of f over I is concave down
*A straight line is neither concave up nor concave down*
Ex. Determine the given intervals in which f(x) = x²+1 is concave up or concave down
x² - 4
Points of Inflection – a point where the graph of a function has a tangent line and where
the concavity changes. (has to change sign)
a) a point on a curve where y′′ is + on one side, - on the other side
b) at such a point y′′ is either 0 or undefined, and y′ has a local max. or min.
Ex. (p.269 book - y = x4 (y´´ = 12x² -- is a zero but no sign change,
Y = x1/3, has a point of inflection at x=0 but y´´ doesn’t
exist there.)
Ex. Determine the points of inflection,discuss the concavity of the graph of f(x)=x4 – 4x³
Second Derivative Test for Local Extrema - used instead of looking for sign change
in f′(x) at critical points.
--Suppose f′′ is continuous on an open interval that contains x = c, then:
1. If f′(c) = 0 and f′′(c)<0, then f has a local max at x = c
2. If f′(c)=0 and f′′(c)>0, then f has a local min. at x =c
3. If f′(c)=0 and f′′(c)=0 then the test fails. The function may have a local min, local max
or neither.
 this test requires that we know f′′ only at c itself and not in interval about c
 test is inconclusive if f′′=0 or f′′ does not exist at x = c – when this happens
use the first derivative test for local extrema
Together f′ and f′′ tell us:
a) The shape of the functions graph
**where the critical points are located and what happens at a critical point
b) Where the function is increasing and where it is decreasing
c) How the curve is bending or turning as defined by concavity
Ex. Sketch the graph of f(x) = x4 – 4x³ + 10
a)
b)
c)
d)
e)
Identify where extrema of f occur
Find the intervals on which f is increasing and decreasing
Find where the graph is concave up and concave down
Sketch the general shape
Plot some specific points (local max., min. points of inflection, intercepts)
Strategy for graphing y =f(x)
#1. Identify the domain of f and any symmetries the curve may have
#2. Find y′ and y′′
#3. Find the critical points of f and identify the function’s behaviors at each one
#4. Find where the curve is increasing and decreasing
#5. Find the points of inflection, if they occur, and determine the concavity of the curve
#6. Identify any asymptotes
#7. Plot key points, such as intercepts and points where y′ and y′′=0, and points of
inflection.
***Know the charts on P. 274
Application: Studying Motion Along a Line
Ex. A particle is moving along a horizontal line with position function of:
S(t) = 2t³ - 14t² + 22t – 5
t≥0
Find the velocity and acceleration and describe the motion of the particle.
Summary of Curve Sketching:
4.6 Optimization Problems
Guidelines for solving Optimization Problems:
#1. Read the problem until you understand it
a) note what is given
b) what is the unknown quantity to be optimized
#2. Draw a picture – label any parts that are important to the problem
#3. Introduce Variables
a) List using relation in the picture and in the problem as an equation or
algebraic expression.
b) Identify the unknown variable
#4. Write an equation for the unknown quantity ---- (Primary Equation)
a) express the unknown as a function of a single variable or in 2 equations in 2
unknowns. (2 equations ---- will require manipulation)
#5. Find the critical points and then test critical points and endpoints of domain in the
unknowns.
Ex. #1 p. 285
What is the smallest perimeter possible for a rectangle whose area is 16 in². What are the
dimensions of the rectangle?
--Use A = l x w
L x w = 16
and P = 2l + 2w
w = 16/l
P = 2l + 2(16/l) ----- 2l + 32/l
** Now find the derivative = 2 - 32/l²
** get l² out of denominator by dividing everything by l² == 2l² - 32
** Set = 0 to find critical points
2l² - 32 = 0
2(l – 4)(l + 4) = 0
L = -4, 4
** must be > 0 because dealing with dimensions so l=4
16 = 4 x w so w = 4 and P = 2(4) + 2(4) = 16
Ex. A manufacturer wants to design an open box having a square base and a surface area
of 108 square inches. What dimensions will produce a box with maximum volume?
Volume = x²h (lxwxh)
Surface Area = area of base + area of all 4 sides
108 = x² + 4xh
Because volume is maximized we want to express V as a function of just 1 variable so
solve for h: x² + 4xh = 108
h = (108 - x²)
4x
V = x² (108-x²)
4x
** now plug h back into original equation
= 108x² - x4
4x
= 27x - x³
4
Now to find maximum value of V – determine a feasible domain – where do the values of
x make sense in this problem?
**know V≥0 and that x can’t be - and that Area of base = x² is at most 108
So feasible Domain = 0≤x≤√108
**to maximize V - find critical numbers so differentiate with respect to x
Dy/dx 27x - x³
= 27 – 3x²
4
4
** now set =0 and find critical pts.
27 – 3x² = 0
4
x²=36
x = -6, 6 (only want +6)
**Now evaluate endpoints and critical points
V(0) = 0 (0,0)
V(6) = 162 – 54 = 108 (6,108) ** so x = 6. h = 3 and x = 6
V(√108) = 0 (√108, 0)
Economics Optimization:
Marginal Revenue dr/dx
Marginal Cost dc/dx
Marginal Profit dp/dx
r(x) = revenue for selling x items
c(x) = cost of producing x items
p(x) = r(x) – c(x) profit from producing and selling x items
if p(x) = r(x) – c(x) has a max value it occurs at production level at which p′(x) = 0
** at a production level yielding max profit – marginal revenue = marginal cost
Ex #43.
You offer a tour service that offers the following:
*$200 per person if 50 people (the min. to book a tour) go on the tour
* For each additional person, up to a max of 80 people total, the rate per person is
reduced by $2
It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people
does it take to maximize your profit?
Let x = the # of people
(50 + x) = the # of people over 50
(200 – 2x) = reduced rate for each person if over 50 book
Cost = 32(50 + x) – 6000
P(x) = r(x) – c(x) so p(x) = (50+x)(200-2x)-32(50 + x) -6000
P(x) = -2x² + 68x + 2400
** Now take derivative and set = 0 to find critical points
-4x + 68 = 0 x = 17
So (50 + 17) would maximize profit = 67 people
4.7 Indeterminate Forms and L’Hopital’s Rule
Suppose we are trying to analyze the behavior of the function f(x) = lnx
x -1
** although f is not defined at x=1, we need to know how f(x) behaves near 1
(in other words – we want to know the value of the limit)
** In computing the limit – we see that the Numerator and Denom. approach 0
and are not defined this leads to:
Indeterminate Form (0/0 form) – occurs when continuous functions f(x) and g(x) are
both 0 at x=a, so lim f(x) cannot be found by substituting x=a, it would result in 0/0
g(x)
x--a
**in our study of limits we found that it we used cancellation, rearrangement of
Terms, or some other algebraic manipulations, it often led to indeterminate 0/0
Form.
** However, if we used the form lim f(x) – f(a) the long form of the derivative
x–a
x—a
it would prevent getting the indeterminate form. This leads to:
L’Hopital’s Rule (First Form)
Suppose that f(a)=g(a) = 0, that f′(a) and g′(a) exist and that g′(a) ≠0 then
Lim f(x) = f′(a)
g′(a)
** treat as 2 separate functions
x ---a g(x)
(show graphical representation)
Ex. a) lim 3x – sinx
x
x---0
b) lim √1 + x - 1
x—0
x
** sometimes, though, after differentiation, the new Numerator and Denominator both
still = 0 at x=a, then we use the stronger form of L’Hopital’s Rule
L’Hopital’s Rule (Stronger Form)
Suppose that f(a) = g(a)=0 and that f and g are differentiable in the open interval, I,
containing a and that g′(x)≠a on I if x≠a then
Lim f(x) = lim f′(x)
x—a g(x)
x—a g′(x)
**basically just keep deriving til no 0 in denomin.
Ex. a) lim √1 + x -1 + x/2
x—0
x²
b) lim x – sinx
x—0
x³
**L’Hopital’s Rule does not apply when either the Numerator or Denominator has
a finite nonzero limit --- Must be indeterminate form of 0/0
Ex. lim 1-cosx
x—0 x + x²
L’Hopital’s Rule with One Sided Limits:
Ex. lim
sin x
+
x—0
x²
lim
sin x
x—0
x²
Indeterminate Forms ∞/∞ - if in the form where both f(x)---∞/-∞ and g(x) --- ∞/-∞,
then lim f′(x) may or may not exist provided that the limit on the right exists
x—a g′(x)
Ex. lim
sec x
x--π/2 1+tan x
b) lim
x- 2x²
x--∞ 3x² + 5x
c) lim ex
x--∞ x²
Indeterminate Form ∞∙0
If lim f(x) = 0 and lim g(x) = ∞/-∞ then it is not clear what the values of lim f(x)g(x)
x—a
x—a
x—a
if any, will be – there becomes a struggle between f and g.
a) if f wins – then the answer = 0
b) if g wins – then the answer may be ∞ or -∞
c) **may be a compromise where the answer is a finite nonzero #
** You deal with this by writing f(x)g(x) as a quotient: fg = f or fg = g
1/g
1/f
(put whichever = 0 in numerator)
Ex. Lim (xsin 1 )
x--∞
x
Steps: 1 evaluate f and g at x--∞
Step 2 Then derive each the n and d separately
Step 3 Solve
Indeterminate Difference
Ex. Lim ( 1 - 1 )
x—0 sinx x
if x—0+ then sin x—0+ and 1/sinx – 1/x = ∞ - ∞
if x—0- then sin x—0- and 1/sinx – 1/x = -∞- -∞=-∞+∞
Steps: to find out what happens to the limit – combine functions:
So:
4.7 Newton’s Method – a technique to approximate the solution to an equation f(x) =0
** It uses tangent lines in place of the graph of y=f(x) near the points where f is 0
GOAL: for estimating a solution of an equation f(x) = 0 is
a) To produce a sequence of approximations that approach the solution
--starts with an initial guess (x0) and improves the guess 1-step at a time.
Refer to graph:
Procedure for Newton’s Method: - let f(c)=0 where f is differentiable on an open
interval Containing c, then to approximate c do the following:
#1. Make an initial estimate, x0 that is close to c
#2. Use the 1st approximation to get a 2nd, then use the 2nd to get a 3rd, and so on using:
Xn+1 = xn - f(xn) if f′(xn) ≠0
f′(xn)
Ex. f(x) = x²-2 we use x0 = 1 as initial guess (graph in calculator)
a) f(x) =x² - 2
f′(x) = 2x
so: using above formula: xn+1 = xn - xn² - 2
2xn
= xn+1 = xn – xn + 1
2 xn
X0=1 so plug in 1 1 - ½ + 1/1 = 1.5
X1 = 1.5
1.5 – .75 + (1/1.5) = 1.41667
X2 = 1.41667
= 1.414216
X3= 1.414216
** if we keep carrying this procedure out you will see it converges
To 1.4142…. where each recursion results in more decimals places
Carried out, but with the same first 5 decimals – approaches a limit
(we know this particular answer is √2 = 1.414213562
Purpose of Newton’s method – it is used to calculate roots b/c they converge so fast.
Ex. Find the x-coordinate of point where curve y = x³ - x crosses horizontal line at y=1
Start by writing f(x) = x³ -x = 1 then solve by getting = to 0
Convergence of Newton’s Method: as in above examples, the approximation approaches
a limit – the sequence x1, x2, x3,…xn is said to converge – therefore
if the limit is c, then it can be shown that c must be a 0 of f.
When Newton’s Method doesn’t yield a convergent sequence:
#1. B/C Newton’s method involves division by f′(xn) – it is clear that the method will
fail if the derivative is 0 for any xn in the sequence
a) Correct this by = overcome it by choosing a different starting value of x
#2 When successive approximations go back and forth between 2 values (iterative
formula.
Conditions for Convergence:
│f(x)f′′(x)│ < 1
│[f′(x)]² │
Ex. f(x) = x²-2
f′(x) = 2x f′′(x) = 2
= │(x²-2)(2)│ = │ 1 - 1 │
│[2x]²
│
│ 2 x² │
What about on interval (1,3) - just plug into above and see if <1 and it will indicate
convergence.
#3. When Newton’s method converges to a root, it may not be the root you have in mind.
Fractal Basins and Newton’s Method
a) finding roots by Newton’s method can be uncertain in the sense that for some
equation ---- the final outcome can be extremely sensitive to the starting values
location
Ex. f(x) = 4x4 – 4x² (look at figure 4.52 in book p. 304
 there are 3 roots you can see from graph
 since there are so many roots you have “basins of attraction”
Fractal Basins – when each root has infinitely many “basins’ of attraction in the complex
plane – each basin has a boundary where complicated patterns repeat without end under
successive manipulations.
Exercise #6 Use Newton’s method to find the negative fourth root of by solving the
equation x4 – 2 = 0. Start with x0 = 1 and find x2
4.8 Antidifferentiation- the process of recovering a function from its known derivative
(known rate of change)
Antiderivative – a function F, is an antiderivative of f on an interval I, if f′(x) = f(x) for
all x in I.
a) Use capital letters to represent an antiderivative of a function, not the
Ex. find the antiderivative of:
h(x) = 2x + cosx
f(x) = 2x
** Now these are not the only functions whose derivative are the above so…
Arbitrary Constant (C) – is put on to represent all forms of the antiderivative that may
have a constant.
a) If F is an antiderivative of f on interval I, then the most general antiderivative
of f on I is: F(x) + C --- where C is an arbitrary constant that represents a
family of functions.
Some common Antiderivative forms:
#1. xn
xn+1 + C
n+1
#2. sinkx
-coskx + C
k
#3. coskx
sinkx + C
k
#4. sec²x
tanx + C
#5. csc²x
-cotx + C
#6. secxtanx
secx + C
#7. cscxcotx
-cscx + C
Ex. Find an antiderivative of f(x) = sin x that satisfies F(0) = 3
Ex. Find antiderivatives of:
a) x5
b) 1
√x
c) sin2x
d) cosx
2
Rules of Antidifferentiation:
original funct.
Antiderivative funct.
#1. Constant Multiple Rule -
kf(x)
kF(x) + C (k is constant)
#2. Negative Rule
-f(x)
-F(x) + C
#3. Sum/Difference Rule
f(x) + g(x)
F(x) + G(x) + C
Ex. Write the antiderivative of √x + 1
√x
Ex. cos πx + πcosx
2
Finding an antiderivative for a function f(x) is the same problem as finding a function
y(x) that satisfies the equation dy = f(x) --- Called a differential Equation
dx
a) Differential Equation – because it is an equation involving an unknown
function, y, that is being differentiated.
**Need a function y(x)- that satisfies the equation – found by taking the
antiderivative of f(x)
** then fix the arbitrary constant arising in the antidifferentiation process by
specifying an initial condition y(x0) = y0
(means the function y(x) has the value y0 when x = x0
Initial Value Problem – the combination of a differential equation and an initial
condition
Ex. Find the curve whose slope at the point (x,y) is 3x² if the curve is required to pass
through the point (1,-1)
Steps: 1
differential equation dy/dx = 3x²
Initial condition y(1) = -1
Step 2: Solve the differential equation (by doing the antiderivative)
Y = x³ + C ---- GENERAL SOLUTION
Step 3: Find the value for the initial condition y(1) = -1
-1 = (1)³ + C
-2 = C
--------PARTICULAR SOLUTION
Step 4: Write the answer y = x³ - 2
Indefinite Integrals - the set of all antiderivatives of f
∫f(x)dx
∫ - integral sign
Integrand = the function f
x = variable of integration
Back to ex. 1
∫2xdx = x² + C
∫cosx dx
∫(2x + cosx)dx
Indefinite Integration done term by term and rewriting of constant of integration:
Ex.
Evaluate ∫( 1 - x² - 1 ) dx
x²